3.91 \(\int \frac {1}{a+b \text {sech}(c+d x)} \, dx\)

Optimal. Leaf size=59 \[ \frac {x}{a}-\frac {2 b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

[Out]

x/a-2*b*arctan((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3783, 2659, 208} \[ \frac {x}{a}-\frac {2 b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x])^(-1),x]

[Out]

x/a - (2*b*ArcTan[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{a+b \text {sech}(c+d x)} \, dx &=\frac {x}{a}-\frac {\int \frac {1}{1+\frac {a \cosh (c+d x)}{b}} \, dx}{a}\\ &=\frac {x}{a}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a d}\\ &=\frac {x}{a}-\frac {2 b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 60, normalized size = 1.02 \[ \frac {\frac {2 b \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {c}{d}+x}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x])^(-1),x]

[Out]

(c/d + x + (2*b*ArcTan[((-a + b)*Tanh[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d))/a

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fricas [A]  time = 0.41, size = 270, normalized size = 4.58 \[ \left [\frac {{\left (a^{2} - b^{2}\right )} d x - \sqrt {-a^{2} + b^{2}} b \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b\right )}}{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) + 2 \, {\left (a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) + a}\right )}{{\left (a^{3} - a b^{2}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x + 2 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)),x, algorithm="fricas")

[Out]

[((a^2 - b^2)*d*x - sqrt(-a^2 + b^2)*b*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) -
a^2 + 2*b^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(-a^2 + b^2)*(a*cosh(d*x + c) + a*sinh(d*x + c
) + b))/(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)*sinh(d*x + c) + a
)))/((a^3 - a*b^2)*d), ((a^2 - b^2)*d*x + 2*sqrt(a^2 - b^2)*b*arctan(-(a*cosh(d*x + c) + a*sinh(d*x + c) + b)/
sqrt(a^2 - b^2)))/((a^3 - a*b^2)*d)]

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giac [A]  time = 0.12, size = 56, normalized size = 0.95 \[ -\frac {\frac {2 \, b \arctan \left (\frac {a e^{\left (d x + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a} - \frac {d x + c}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)),x, algorithm="giac")

[Out]

-(2*b*arctan((a*e^(d*x + c) + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*a) - (d*x + c)/a)/d

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maple [A]  time = 0.21, size = 88, normalized size = 1.49 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sech(d*x+c)),x)

[Out]

-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)-2/d*b/a/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1
/2))+1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 0.40, size = 131, normalized size = 2.22 \[ \frac {x}{a}+\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2}-\frac {2\,b\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^2\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {b-a}}-\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2}+\frac {2\,b\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^2\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {b-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cosh(c + d*x)),x)

[Out]

x/a + (b*log((2*b*exp(c + d*x))/a^2 - (2*b*(a + b*exp(c + d*x)))/(a^2*(a + b)^(1/2)*(b - a)^(1/2))))/(a*d*(a +
b)^(1/2)*(b - a)^(1/2)) - (b*log((2*b*exp(c + d*x))/a^2 + (2*b*(a + b*exp(c + d*x)))/(a^2*(a + b)^(1/2)*(b -
a)^(1/2))))/(a*d*(a + b)^(1/2)*(b - a)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \operatorname {sech}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c)),x)

[Out]

Integral(1/(a + b*sech(c + d*x)), x)

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